Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+9y &= -6 \\ 8x+3y &= 2\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = -3y+2$ Divide both sides by $8$ to isolate $x$ $x = {-\dfrac{3}{8}y + \dfrac{1}{4}}$ Substitute this expression for $x$ in the first equation. $6({-\dfrac{3}{8}y + \dfrac{1}{4}}) + 9y = -6$ $-\dfrac{9}{4}y + \dfrac{3}{2} + 9y = -6$ Simplify by combining terms, then solve for $y$ $\dfrac{27}{4}y + \dfrac{3}{2} = -6$ $\dfrac{27}{4}y = -\dfrac{15}{2}$ $y = -\dfrac{10}{9}$ Substitute $-\dfrac{10}{9}$ for $y$ in the top equation. $6x+9( -\dfrac{10}{9}) = -6$ $6x-10 = -6$ $6x = 4$ $x = \dfrac{2}{3}$ The solution is $\enspace x = \dfrac{2}{3}, \enspace y = -\dfrac{10}{9}$.